\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 19, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/19\hfil Third-order three-point nonlocal problems]
{Third-order nonlocal problems with sign-changing nonlinearity on time scales}

\author[D. R. Anderson, C. C. Tisdell\hfil EJDE-2007/19\hfilneg]
{Douglas R. Anderson, Christopher C. Tisdell}  % in alphabetical order

\address{Douglas R. Anderson \newline
Department of Mathematics and Computer Science,
Concordia College, Moorhead, MN 56562, USA}
\email{andersod@cord.edu}

\address{Christopher C. Tisdell \newline
School of Mathematics and Statistics, University of New South Wales,
Sydney, UNSW 2052, Australia}
\email{cct@unsw.edu.au}

\thanks{Submitted October 31, 2006. Published January 27, 2007.}
\thanks{Supported by grant DP0450752 from the Australian
Research Council's Discovery Projects.}
\subjclass[2000]{34B18, 34B27, 34B10, 39A10}
\keywords{Boundary value problem; time scale; three-point;
 Green's function}

\begin{abstract}
 We are concerned with the existence and form of positive
 solutions to a nonlinear third-order three-point nonlocal
 boundary-value problem on general time scales. Using Green's functions,
 we prove the existence of at least one positive solution
 using the Guo-Krasnoselskii fixed point theorem.
 Due to the fact that the nonlinearity is allowed to change
 sign in our formulation, and the novelty of the boundary conditions,
 these results are new for discrete, continuous, quantum and arbitrary
 time scales.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}

\section{Statement of the problem}

We will develop an interval of $\lambda$ values whereby a positive
solution exists for the following nonlinear, third-order,
three-point, nonlocal boundary-value problem on arbitrary time
scales
\begin{gather}
   (px^{\Delta\Delta})^\nabla(t)=\lambda f(t,x(t)),
 \quad t\in[t_1,t_3]_\mathbb{T},  \label{bvp} \\
   \alpha x(\rho(t_1))-\beta x^\Delta(\rho(t_1))
 =\int_{\xi_1}^{\xi_2} a(t)x(t)\nabla t,  \nonumber  \\
    x^\Delta(t_2)=0, \quad (px^{\Delta\Delta})(t_3)
=\int_{\eta_1}^{\eta_2} b(t)(px^{\Delta\Delta})(t)\nabla t, \label{bvpbc}
\end{gather}
where: $p$ is a left-dense continuous, real-valued function on
$\mathbb{T}$  with $p>0$; $\lambda>0$ is a real scalar;
\begin{itemize}
 \item[(H1)] the real scalars $\alpha,\beta>0$ and the three boundary points satisfy $t_1<t_2<t_3\in\mathbb{T}$ such that
      $$ 0<\int_{\rho(t_1)}^{\sigma^2(t_3)}\int_u^{t_2}\frac{\Delta r}{p(r)}\Delta u
        +\frac{\beta}{\alpha}\int_{\rho(t_1)}^{t_2}\frac{\Delta r}{p(r)} < \infty; $$
 \item[(H2)] the points $\xi_i,\eta_i\in\mathbb{T}$ satisfy
      $$ \rho(t_1)<\xi_1<\xi_2<t_2, \quad \rho(t_1)\le \eta_1<\eta_2\le t_3; $$
 \item[(H3)] the left-dense continuous real-valued functions on $\mathbb{T}$ satisfy $a,b\ge 0$ with
      $$ 0<\int_{\xi_1}^{\xi_2}a(t)\nabla t<\alpha \quad\text{and}\quad 0<\int_{\eta_1}^{\eta_2}b(t)\nabla t<1; $$
      \item[(H4)] the continuous function $f:[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}\times[0,\infty)\to(-\infty,\infty)$
      is such that
      $$ \lim_{x\to +\infty}\frac{f(t,x)}{x}\stackrel{\text{unif}}=+\infty, \quad t\in[\xi_1,t_2]_\mathbb{T}; $$
 \item[(H5)] there exist left-dense continuous functions $y,z:[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}\to(0,\infty)$
      and a continuous function $h:[0,\infty)\to(0,\infty)$ such that
      $$ -y(t)\le f(t,x)\le z(t)h(x), \quad  0 < \int_{\rho(t_1)}^{t_3}(z(s)+y(s))\nabla s<\infty. $$
\end{itemize}

Third-order differential equations, though less common in
applications  than even-order problems, nevertheless do appear,
for example in the study of quantum fluids; see Gamba and
J\"{u}ngel \cite{gamba}.  Here we approach third-order problems on
general time scales, namely on any nonempty closed subset of the
real line, to include the discrete, continuous, and quantum
calculus as special cases. Boundary value problems on time scales
that utilize both delta and nabla derivatives, such as the one
here, were first introduced by Atici and Guseinov \cite{ag}.
Three-point and right-focal boundary value problems, in both the
continuous and discrete cases, have been addressed in
\cite{and,and2,and3,ad}, by Eloe and McKelvey \cite{em}, and
recently by Graef and Yang \cite{gy,gy2}, Sun \cite{sun}, and Wong
\cite{wong}.  For more on existence of solutions to boundary value
problems, see \cite[Chapters 4 and 6-9]{bp}, Davis, Erbe, and
Henderson \cite{deh}, Erbe and Wang \cite{ew}, the text by Guo and
Lakshmikantham \cite{gu}, Henderson \cite{hen}, Henderson and
Thompson \cite{ht}, Lan \cite{lan,lan2}, Ma and Thompson
\cite{mt}, and Zhang and Liu \cite{zl}. Problem \eqref{b1},
\eqref{b2} is an extension of the continuous and discrete
discussions of third-order right-focal boundary value problems to
time scales, and by the addition of the nonhomogeneous nonlocal
boundary conditions and the allowance of sign changes in the
nonlinearity $f$, problem \eqref{bvp}, \eqref{bvpbc} is introduced
for the first time on any time scale, including $\mathbb{R}$, $\mathbb{Z}$, and
the quantum time scale. One could also consider a third-order
problem with derivatives in the order of nabla, nabla, delta, but
the results would be similar; other permutations of nablas and/or
deltas lead to a Green function that is less easy to calculate.

Clearly there are other approaches to the existence of positive
solutions  for dynamic equations on time scales than those
featured in this work;  for alternative approaches to the
existence of solutions and multiple solutions to dynamic equations
on time scales, consult, for example, Bohner and Luo \cite{bl},
Henderson \cite{hen2}, Ma, Du, and Ge \cite{mdg}, and Tisdell,
Dr\'{a}bek, and Henderson \cite{tisdell}. Underlying our
technique, however, will be Green's function for the homogeneous,
third-order, three-point boundary-value problem
\begin{gather}
  (px^{\Delta\Delta})^\nabla(t)=0, \quad t\in[t_1,t_3]_\mathbb{T}, \label{b1}\\
  \alpha x(\rho(t_1))-\beta x^\Delta(\rho(t_1))=x^\Delta(t_2)=(px^{\Delta\Delta})(t_3)=0.\label{b2}
\end{gather}
Green's function for \eqref{b1}, \eqref{b2} will be  defined on
$[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$, nonnegative on
$[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$ contingent on the distance between
boundary points, nondecreasing on $[\rho(t_1),t_2]_\mathbb{T}$, and
nonincreasing on $[t_2,\sigma^2(t_3)]_\mathbb{T}$, as will be shown in the
following lemmas.


\begin{lemma} \label{lem1.1}
Green's function corresponding to the problem \eqref{b1},
\eqref{b2}  is given by
\begin{equation}\label{greenf}
\begin{aligned}
&G(t,s)\\
&=\begin{cases}
    s\in [\rho(t_1),t_2]_\mathbb{T} &:
        \begin{cases}
            \int_{\rho(t_1)}^t\int_u^{s}\frac{\Delta r}{p(r)}\Delta u
+ \frac{\beta}{\alpha}\int_{\rho(t_1)}^{s}\frac{\Delta r}{p(r)}   &:t<s \\[4pt]
      \int_{\rho(t_1)}^s\int_u^{s}\frac{\Delta r}{p(r)}\Delta u + \frac{\beta}{\alpha}\int_{\rho(t_1)}^{s}\frac{\Delta r}{p(r)} &:t\geq s
        \end{cases}\\[14pt]
     s\in [t_2,\sigma^2(t_3)]_\mathbb{T} &:
        \begin{cases}
         \int_{\rho(t_1)}^t\int_u^{t_2}\frac{\Delta r}{p(r)}\Delta u
+ \frac{\beta}{\alpha}\int_{\rho(t_1)}^{t_2}\frac{\Delta r}{p(r)} &:t<s \\[4pt]
         \int_{\rho(t_1)}^t\int_u^{t_2}\frac{\Delta r}{p(r)}\Delta u + \int_{s}^t\int_{s}^{u}\frac{\Delta r}{p(r)}\Delta u + \frac{\beta}{\alpha}\int_{\rho(t_1)}^{t_2}\frac{\Delta r}{p(r)} &:t\geq s.
        \end{cases} \\
    \end{cases}
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
We follow the approach given, for example, in Kelley and  Peterson
\cite[Chapter 5]{kp}.  As the Cauchy function $y(\cdot,s)$
satisfies the homogeneous time-scale initial-value problem
$$
(py^{\Delta\Delta}(\cdot,s))^\nabla(t)=0, \quad y(s,s)=0,
\quad y^\Delta(s,s)=0, \quad y^{\Delta\Delta}(\rho(s),s)=1/p(\rho(s))
$$
it is easy to verify that
$y(t,s)=\int_{s}^t\int_{s}^{u}\frac{\Delta r}{p(r)}\Delta u$.
Thus the Green function takes the form
$$  G(t,s)=
    \begin{cases}
        s\in [\rho(t_1),t_2]_\mathbb{T} &:
            \begin{cases}
                u_1(t,s)    &:t<s \\
        v_1(t,s)    &:t\geq s
            \end{cases}\\[12pt]
        s\in [t_2,\sigma^2(t_3)]_\mathbb{T} &:
            \begin{cases}
                u_2(t,s)    &:t<s \\
        v_2(t,s)    &:t\geq s,
            \end{cases}
    \end{cases} $$
where $u_i(t,s)+y(t,s)=v_i(t,s)$ and $v_i(\cdot,s)$ satisfy $(px^{\Delta\Delta})^\nabla(t)=0$, for $i=1,2$. Let $s\in [\rho(t_1),t_2]_\mathbb{T}$. Then the boundary conditions are $\alpha u_1(\rho(t_1),s)-\beta u_1^\Delta(\rho(t_1),s)=0$ for $t<s$ and $v_1^\Delta(t_2,s)=(pv_1^{\Delta\Delta}(\cdot,s))(t_3)=0$ for $s\le t$.  Solving for $v_1$, we see that $v_1(t,s)=k(s)$ for some function $k$.  Since $u_1=v_1-y$, $\alpha u_1(\rho(t_1),s)=\beta u_1^\Delta(\rho(t_1),s)$ for these $s$ implies that
$$ k(s)=\int_{\rho(t_1)}^s\int_u^{s}\frac{\Delta r}{p(r)}\Delta u + \frac{\beta}{\alpha}\int_{\rho(t_1)}^{s}\frac{\Delta r}{p(r)}. $$
Thus for $s\in [\rho(t_1),t_2]_\mathbb{T}$,
$$ G(t,s) =
            \begin{cases}
                \int_{\rho(t_1)}^t\int_u^{s}\frac{\Delta r}{p(r)}\Delta u   + \frac{\beta}{\alpha}\int_{\rho(t_1)}^{s}\frac{\Delta r}{p(r)} &:t<s \\
        \int_{\rho(t_1)}^s\int_u^{s}\frac{\Delta r}{p(r)}\Delta u   + \frac{\beta}{\alpha}\int_{\rho(t_1)}^{s}\frac{\Delta r}{p(r)} &:t\geq s.
            \end{cases}
$$
Now let $s\in [t_2,\sigma^2(t_3)]_\mathbb{T}$, so that the boundary
conditions  are $\alpha u_2(\rho(t_1),s)-\beta
u_2^\Delta(\rho(t_1),s)=u_2^\Delta(t_2,s)=0$ for $t<s$ and
$(pv_2^{\Delta\Delta}(\cdot,s))(t_3)=0$ for $s\le t$.  Clearly
$$
u_2(t,s)=-q(s)\Big(\int_{\rho(t_1)}^t\int_u^{t_2}\frac{\Delta
r}{p(r)}\Delta u + \frac{\beta}{\alpha}
\int_{\rho(t_1)}^{t_2}\frac{\Delta r}{p(r)}\Big)
 $$
for some function $q$. Using the fact that $v_2=u_2+y$ and the
remaining boundary condition yields $q(s)\equiv -1$ and
$$
G(t,s) =
            \begin{cases}
                \int_{\rho(t_1)}^t\int_u^{t_2}\frac{\Delta r}{p(r)}\Delta
+ \frac{\beta}{\alpha} \int_{\rho(t_1)}^{t_2}\frac{\Delta r}{p(r)} &:t<s \\[4pt]
                \int_{\rho(t_1)}^t\int_u^{t_2}\frac{\Delta r}{p(r)}\Delta
+ \int_{s}^t\int_{s}^{u}\frac{\Delta r}{p(r)}\Delta u + \frac{\beta}{\alpha} \int_{\rho(t_1)}^{t_2}\frac{\Delta r}{p(r)} &:t\geq s
            \end{cases}
$$
for $s\in [t_2,\sigma^2(t_3)]_\mathbb{T}$.
\end{proof}


\begin{remark} \label{rmk1.2} \rm
As in \eqref{greenf} and the proof above, throughout the rest of
the paper  we take
\begin{equation}\label{u2}
 u_2(t):=\int_{\rho(t_1)}^{t}\int_u^{t_2}\frac{\Delta r}{p(r)}\Delta u
+ \frac{\beta}{\alpha}\int_{\rho(t_1)}^{t_2}\frac{\Delta r}{p(r)}.
\end{equation}
\end{remark}

\begin{lemma}\label{lemma13}
Green's function \eqref{greenf} corresponding to the problem
 \eqref{b1}, \eqref{b2} satisfies
$$
0 < G(t,s) \le G(t_2,s) \le u_2(t_2) $$
for
$(t,s)\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}\times(\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$
if and only if {\rm (H1)} holds; that is, $u_2(\sigma^2(t_3))>0$.
\end{lemma}

\begin{proof}
Fix $s\in(\rho(t_1),t_2]_\mathbb{T}$. Then
$u_1(\rho(t_1),s)=\frac{\beta}{\alpha}\int_{\rho(t_1)}^s\frac{\Delta
r}{p(r)}>0$ and $u_1(\cdot,s)$ is increasing, so that
$0<u_1(t,s)\le u_1(s,s)$ for $t\in[\rho(t_1),s)_\mathbb{T}$. But
$u_1(s,s)\equiv v_1(t,s)$ for $t\in[t_2,\sigma^2(t_3)]_\mathbb{T}$.  It
follows that $G(t_2,s)\geq G(t,s)>0$ for $s\in(\rho(t_1),t_2]_\mathbb{T}$
and $t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$.  Now fix
$s\in[t_2,\sigma^2(t_3)]_\mathbb{T}$. The branch $u_2$ is positive at
$\rho(t_1)$, increases until $t_2$, and then decreases until $s$.
We then switch to branch $v_2$, which continues to decrease, so
that $v_2(t,s)\ge v_2(\sigma^2(t_3),s)$.  As a function of $s$,
$v_2(\sigma^2(t_3),s)$ is also decreasing, whence $v_2(t,s)\ge
v_2(\sigma^2(t_3),\sigma^2(t_3))=u_2(\sigma^2(t_3))$ for $u_2$
given in \eqref{u2}. Thus $G(t_2,s)\ge G(t,s)$ for
$s\in[t_2,\sigma^2(t_3)]_\mathbb{T}$ and
$t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$ as well, and $G(t,s)>0$ for
$s\in[t_2,\sigma^2(t_3)]_\mathbb{T}$ and
$t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$ if and only if (H1) holds.
\end{proof}

\begin{remark}\label{rmk1.4} \rm
If $\mathbb{T}=\mathbb{Z}$, $\alpha=1$, $\beta=0$, and $p(t)\equiv 1$, then the
necessary and sufficient condition for the Green function to be
positive is $t_2-t_1-1 \ge t_3-t_2$; see \cite{and2}.
\end{remark}

\begin{lemma}\label{lemma15}
Assume {\rm (H1)}. For any
$(t,s)\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}\times(\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$,
the Green function \eqref{greenf} corresponding to the problem
 \eqref{b1}, \eqref{b2} satisfies, using \eqref{u2},
$$
\frac{u_2(t)}{u_2(t_2)} \le \frac{G(t,s)}{G(t_2,s)} \le 1.
 $$
\end{lemma}

\begin{proof}
The right-hand inequality follows from the previous lemma.
For the left-hand inequality, we proceed by analyzing branches of the
Green function \eqref{greenf}. For fixed
$t\in[\rho(t_1),s)_\mathbb{T}$ and $s\in(t,t_2]_\mathbb{T}$,
$$
\frac{G(t,s)}{G(t_2,s)} = \frac{\int_{\rho(t_1)}^t\int_u^{s}
\frac{\Delta r}{p(r)}\Delta u+\frac{\beta}{\alpha}\int_{\rho(t_1)}^{s}
\frac{\Delta r}{p(r)}}{\int_{\rho(t_1)}^{s}\int_u^{s}\frac{\Delta r}{p(r)}
\Delta u+\frac{\beta}{\alpha}\int_{\rho(t_1)}^{s}\frac{\Delta r}{p(r)}}
=:\phi(s).
$$
Then
\begin{align*}
&\phi^{\nabla}(s)\\
&=\frac{(t-s)\left(\int_{\rho(t_1)}^t\int_u^s
\frac{\Delta r}{p(r)}\Delta u+\frac{\beta}{\alpha}\int_{\rho(t_1)}^{s}
\frac{\Delta r}{p(r)} \right)+ \left(t-\rho(t_1)+\frac{\beta}{\alpha}\right)
\int_t^s\int_u^s\frac{\Delta r}{p(r)}\Delta u}{p^{\rho}(s)
\left(\int_{\rho(t_1)}^{s}\int_u^{s}\frac{\Delta r}{p(r)}\Delta u
+\frac{\beta}{\alpha}\int_{\rho(t_1)}^{s}\frac{\Delta r}{p(r)}\right)
\left(\int_{\rho(t_1)}^{\rho(s)}\int_u^{\rho(s)}\frac{\Delta r}{p(r)}
\Delta u+\frac{\beta}{\alpha}\int_{\rho(t_1)}^{\rho(s)}\frac{\Delta r}{p(r)}
\right)}.
\end{align*}
The denominator of $\phi^\nabla$ is clearly positive, so consider the numerator,
\begin{align*}
  \psi(s)&:= \psi_1(s)+\psi_2(s) \\
         &:= (t-s)\int_{\rho(t_1)}^t\int_u^s\frac{\Delta r}{p(r)}\Delta u
+ \left(t-\rho(t_1)\right)\int_t^s\int_u^s\frac{\Delta r}{p(r)}\Delta u \\
 &\quad +\frac{\beta}{\alpha}\Big((t-s)\int_{\rho(t_1)}^{s}
\frac{\Delta r}{p(r)}+\int_t^s\int_u^s\frac{\Delta r}{p(r)} \Delta
u \Big).
\end{align*}
Note that $\psi(t)=0$; the first part satisfies
$$
\psi_1^\nabla(s)=\frac{(t-\rho(t_1))\nu(s)}{p^\rho(s)}
-\int_{\rho(t_1)}^t\int_u^s\frac{\Delta r}{p(r)}\Delta u
=-\int_{\rho(t_1)}^t\int_u^{\rho(s)}\frac{\Delta r}{p(r)}\Delta u \le 0
$$
for $s\in(t,t_2]_\mathbb{T}$, while the second part satisfies
$$
\psi_2(s)\le \frac{\beta}{\alpha}\Big((t-s)\int_{\rho(t_1)}^{s}
\frac{\Delta r}{p(r)}+\int_t^s\int_t^s\frac{\Delta r}{p(r)}\Delta u \Big)
=\frac{\beta}{\alpha}(t-s)\int_{\rho(t_1)}^{t}\frac{\Delta r}{p(r)} \le 0.
$$
Therefore, the whole numerator satisfies $\psi(s)\le 0$, so that
$\phi^\nabla\le 0$ and $\phi$ is nonincreasing as a function of
$s$.  Thus $\phi(s)\ge \phi(t_2)$ for $s\in(t,t_2]$; in other
words,
$$
\frac{G(t,s)}{G(t_2,s)} \ge \frac{G(t,t_2)}{G(t_2,t_2)}
= \frac{u_2(t)}{u_2(t_2)}.
$$
For $s\in[\rho(t_1),t_2]_\mathbb{T}$ and $t\in[s,\sigma^2(t_3)]_\mathbb{T}$,
$$
\frac{G(t,s)}{G(t_2,s)}\equiv 1 \ge \frac{u_2(t)}{u_2(t_2)}.
$$
If $s\in[t_2,\sigma^2(t_3)]_\mathbb{T}$ and $t\in[\rho(t_1),s)_\mathbb{T}$, then
$$
 \frac{G(t,s)}{G(t_2,s)} = \frac{u_2(t)}{u_2(t_2)}.
$$
Finally, if $s\in[t_2,\sigma^2(t_3)]_\mathbb{T}$ and $t\in[s,\sigma^2(t_3)]_\mathbb{T}$, then
$$
\frac{G(t,s)}{G(t_2,s)}=\frac{u_2(t)+\int_s^t\int_{s}^u
\frac{\Delta r}{p(r)}\Delta u}{u_2(t_2)} \ge \frac{u_2(t)}{u_2(t_2)}.
$$
\end{proof}

\section{Exploring the nonlocal problem}

In this section we turn our attention to the problem
\begin{equation}\label{bvp2}
 (px^{\Delta\Delta})^\nabla(t)=\lambda y(t),  \quad t\in[t_1,t_3]_\mathbb{T},
\end{equation}
with nonlocal boundary conditions \eqref{bvpbc}, where $y$ is as
in (H5), and $\lambda>0$. Assume (H2) and (H3), and and use
\eqref{u2} to define
\begin{equation}\label{D}
  D:=u_2(t_2)\Big(1-\int_{\eta_1}^{\eta_2}b(t)\nabla t\Big)
\Big(\int_{\xi_1}^{\xi_2} a(t)\nabla t-\alpha\Big) < 0.
\end{equation}


\begin{lemma}
Assume {\rm (H1)} through {\rm (H5)}. Then the nonhomogeneous
dynamic equation \eqref{bvp2} with boundary conditions
\eqref{bvpbc} has a unique solution $x^*$, where for $t \in
[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$,
\begin{equation}\label{form}
  x^*(t)=\lambda\Big(\int_{\rho(t_1)}^{t_3} G(t,s)y(s)
\nabla s+A(y)u_2(t)+B(y)\left(u_2(t_2)-u_2(t)\right)\Big)
\end{equation}
holds, where: $G(t,s)$ is the Green function \eqref{greenf} of the
boundary-value problem
\eqref{b1}, \eqref{b2}; and the functionals $A$ and $B$ are defined
using \eqref{u2} by
\begin{equation}\label{A}
 A(y):=\frac{1}{D}\begin{vmatrix}
         \int_{\eta_1}^{\eta_2}b(t)\nabla t-1 & \int_{\xi_1}^{\xi_2}a(t)u_2(t)\nabla t+u_2(t_2)\left(\alpha-\int_{\xi_1}^{\xi_2}a(t)\nabla t\right) \\
         \int_{\eta_1}^{\eta_2}b(t)\left(\int_{t}^{t_3}y(s)\nabla s\right)\nabla t &  \int_{\xi_1}^{\xi_2}a(t)\left(\int_{\rho(t_1)}^{t_3}G(t,s)y(s)\nabla s\right)\nabla t  \end{vmatrix},
\end{equation}
\begin{equation}\label{B}
 B(y):=\frac{1}{D}\begin{vmatrix}
         \int_{\eta_1}^{\eta_2}b(t)\nabla t-1 & \int_{\xi_1}^{\xi_2}a(t)u_2(t)\nabla t \\
         \int_{\eta_1}^{\eta_2}b(t)\left(\int_{t}^{t_3}y(s)\nabla s\right)\nabla t &  \int_{\xi_1}^{\xi_2}a(t)\left(\int_{\rho(t_1)}^{t_3}G(t,s)y(s)\nabla s\right)\nabla t  \end{vmatrix}.
\end{equation}
\end{lemma}

\begin{proof}
For $y$ as in (H5), we show that the function $x^*$ given in \eqref{form} is a solution of \eqref{bvp2} with conditions \eqref{bvpbc} only if $A(y)$ and $B(y)$ are given by \eqref{A} and \eqref{B}, respectively.  If $x^*$ is a solution of
\eqref{bvp2}, \eqref{bvpbc}, then
$$ x^*(t)=\lambda\int_{\rho(t_1)}^t G(t,s)y(s)\nabla s + \lambda\int_{t}^{t_3} G(t,s)y(s)\nabla s + Au_2(t) + B(u_2(t_2)-u_2(t)) $$
for some constants $A$ and $B$.  Taking the delta derivative with respect to $t$ yields
$$
x^{*\Delta}(t) = \lambda\int_{\rho(t_1)}^t G^\Delta(t,s)y(s)\nabla s
        + \lambda\int_{t}^{t_3} G^\Delta(t,s)y(s)\nabla s
+ A\int_t^{t_2}\frac{\Delta r}{p(r)} - B\int_t^{t_2}\frac{\Delta r}{p(r)};
$$
since $p$ times the delta derivative of this expression is
$$
(px^{*\Delta\Delta})(t) = -\lambda \int_t^{t_3}y(s)\nabla s-A+B,
$$
we see that \eqref{bvp2} holds.  It is also clear that $x^{*\Delta}(t_2)=0$ is satisfied. To meet the other two boundary conditions in \eqref{bvpbc}, we must have at $\rho(t_1)$ that
\begin{equation}\label{311}
 \alpha Bu_2(t_2)= \int_{\xi_1}^{\xi_2}a(t)
\Big(\lambda\int_{\rho(t_1)}^{t_3}G(t,s)y(s)\nabla s+Au_2(t)+ B(u_2(t_2)
-u_2(t))\Big)\nabla t,
\end{equation}
while at $t_3$ we have
\begin{equation}\label{312}
   -A+B = \int_{\eta_1}^{\eta_2}b(t)
\Big(-\lambda\int_{t}^{t_3}y(s)\nabla s-A+B\Big)\nabla t.
\end{equation}
Combining \eqref{311} and \eqref{312}, we arrive at the system of equations
\begin{align*}
& A\int_{\xi_1}^{\xi_2}a(t)u_2(t)\nabla t
+ B\Big[\int_{\xi_1}^{\xi_2}a(t)\left(u_2(t_2)-u_2(t)\right)\nabla t
-\alpha u_2(t_2)\Big] \\
&=  -\lambda\int_{\xi_1}^{\xi_2}a(t)
\Big(\int_{\rho(t_1)}^{t_3}G(t,s)y(s)\nabla s\Big)\nabla t
\end{align*}
and
$$
A\Big[\int_{\eta_1}^{\eta_2}b(t)\nabla t-1\Big]
+ B\Big[1-\int_{\eta_1}^{\eta_2}b(t)\nabla t\Big]
= -\lambda\int_{\eta_1}^{\eta_2}b(t)\Big(\int_{t}^{t_3}y(s)\nabla s\Big)
\nabla t.
$$
The determinant of the coefficients of $A$ and $B$ is $D$, given by \eqref{D},
which is negative, and by elementary linear algebra we verify \eqref{A} and
 \eqref{B} with $\lambda$ factored out. Also note that $A(y)>B(y)>0$
since $D<0$ and
$$
A(y)-B(y)=\frac {\int_{\eta_1}^{\eta_2}b(t)\Big(\int_{t}^{t_3}y(s)
\nabla s\Big)\nabla t}{1-\int_{\eta_1}^{\eta_2}b(t)\nabla t}.
$$
\end{proof}

\begin{corollary} \label{coro2.2}
Assume {\rm (H1)} through {\rm (H5)}. Then the unique solution
$x^*$ as in \eqref{form} of the problem \eqref{bvp2},
\eqref{bvpbc} satisfies $x^*(t)\ge 0$ for
$t\in[\rho(t_1),\sigma^2(t_3)]$.
\end{corollary}

\begin{proof}
 From Lemma \ref{lemma13} we know that on
the Green function \eqref{greenf} satisfies $G(t,s)\ge 0$.
Assumption (H3) applied to \eqref{A} and \eqref{B} imply that $A(y)>B(y)>0$.
\end{proof}

\begin{lemma}\label{lemma23}
Assume {\rm (H1)} through {\rm (H5)}. Then the unique solution
$x^*$ as in \eqref{form} of the problem \eqref{bvp2},
\eqref{bvpbc} satisfies
$$ \theta\|x^*\| \le x^*(t) \le \lambda \theta \Theta \quad\text{on}\quad [\rho(t_1),\sigma^2(t_3)]_\mathbb{T}, $$
where using \eqref{u2} we take
\begin{equation}\label{theta}
 \theta:=\min\left\{\frac{u_2(\rho(t_1))}{u_2(t_2)},\frac{u_2(\sigma^2(t_3))}{u_2(t_2)}\right\}\in(0,1), \quad \|x^*\|:=\max_{t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}}x^*(t)=x^*(t_2),
\end{equation}
and
\begin{equation}\label{Theta}
 \Theta:=\frac{1}{\theta}u_2(t_2)\left(1 + \bar{A}\right)\int_{\rho(t_1)}^{t_3}y(s)\nabla s
\end{equation}
for
$$ \bar{A}:=\frac{1}{D}\begin{vmatrix}
         \int_{\eta_1}^{\eta_2}b(t)\nabla t-1 & \int_{\xi_1}^{\xi_2}a(t)u_2(t)\nabla t+u_2(t_2)\left(\alpha-\int_{\xi_1}^{\xi_2}a(t)\nabla t\right) \\
         \int_{\eta_1}^{\eta_2}b(t)\nabla t &  u_2(t_2)\int_{\xi_1}^{\xi_2}a(t)\nabla t \end{vmatrix}. $$
\end{lemma}

\begin{proof}
>From previous work, it is clear that for all $t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$,
$$
x^*(t) \le x^*(t_2)
=\lambda\Big(\int_{\rho(t_1)}^{t_3}G(t_2,s)y(s)\nabla s+A(y)u_2(t_2) \Big).
$$
For $t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$, from Lemma \ref{lemma13} and
Lemma \ref{lemma15}, the Green function \eqref{greenf} satisfies
$$
 \frac{G(t,s)}{G(t_2,s)} \ge \frac{u_2(t)}{u_2(t_2)}
\ge \min\Big\{\frac{u_2(\rho(t_1))}{u_2(t_2)},
\frac{u_2(\sigma^2(t_3))}{u_2(t_2)}\Big\} = \theta\in(0,1)
$$
by (H1) and \eqref{u2}, and
\begin{align*}
 & x^*(t) \\
&=  \lambda\Big(\int_{\rho(t_1)}^{t_3}
 \frac{G(t,s)}{G(t_2,s)}G(t_2,s)y(s)\nabla s+A(y)\frac{u_2(t)}{u_2(t_2)}
u_2(t_2)+B(y)(u_2(t_2)-u_2(t))\Big) \\
      &\ge \lambda\Big(\int_{\rho(t_1)}^{t_3} \theta G(t_2,s)y(s)
\nabla s+A(y)\theta u_2(t_2)\Big) \\
      & =  \theta\|x^*\|.
\end{align*}
Consequently, $\theta\|x^*\| \le x^*(t)$ for all
$t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$. For $D$ in \eqref{D} and
$A(y)$ in \eqref{A},
\begin{equation}
\begin{aligned}
& A(y)\\
& =  \frac{1}{D}\begin{vmatrix}
         \int_{\eta_1}^{\eta_2}b(t)\nabla t-1
   & \int_{\xi_1}^{\xi_2}a(t)u_2(t)\nabla t+u_2(t_2)
     \left(\alpha-\int_{\xi_1}^{\xi_2}a(t)\nabla t\right) \\
   \int_{\eta_1}^{\eta_2}b(t)\left(\int_{t}^{t_3}y(s)\nabla s\right)\nabla t
   &  \int_{\xi_1}^{\xi_2}a(t)\left(\int_{\rho(t_1)}^{t_3}G(t,s)y(s)
      \nabla s\right)\nabla t  \end{vmatrix} \nonumber \\
&\le \frac{1}{D}\begin{vmatrix}
         \int_{\eta_1}^{\eta_2}b(t)\nabla t-1
    & \int_{\xi_1}^{\xi_2}a(t)u_2(t)\nabla t+u_2(t_2)
      \left(\alpha-\int_{\xi_1}^{\xi_2}a(t)\nabla t\right) \\
      \int_{\eta_1}^{\eta_2}b(t)\nabla t &  u_2(t_2)
      \int_{\xi_1}^{\xi_2}a(t)\nabla t \end{vmatrix}
      \int_{\rho(t_1)}^{t_3}y(s)\nabla s \nonumber \\
& =  \bar{A}\int_{\rho(t_1)}^{t_3}y(s)\nabla s < \infty.
\end{aligned} \label{AAbar}
\end{equation}
As a result, for $t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$,
\begin{align*}
  x^*(t) &\le \lambda
\Big(\int_{\rho(t_1)}^{t_3}G(t_2,s)y(s)\nabla s+A(y)u_2(t_2) \Big) \\
      &\le \lambda u_2(t_2)(1+\bar{A})\int_{\rho(t_1)}^{t_3} y(s)\nabla s \\
      &\le \lambda\theta\Theta
\end{align*}
using \eqref{theta}, \eqref{Theta}, and \eqref{AAbar}.
\end{proof}

\section{An existence result on cones}

Let $\mathcal{B}$ denote the Banach space $C[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$
with the norm
$$
\|x\|=\sup_{t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}}|x(t)|.
$$
Define the cone $\mathcal{P}\subset\mathcal{B}$ by
$$
 \mathcal{P}=\{x\in\mathcal{B}: x(t)\ge \theta\|x\| \;\text{on}\;
[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}\},
$$
where $\theta$ is given in \eqref{theta}. Consider the related boundary-value
problem
\begin{gather*}
  (px^{\Delta\Delta})^\nabla(t)=\lambda f^*(t,x(t)),  \quad
   t\in[t_1,t_3]_\mathbb{T},  \\
 \alpha x(\rho(t_1))-\beta x^\Delta(\rho(t_1))=\int_{\xi_1}^{\xi_2}
   a(t)x(t)\nabla t,  \\
  x^\Delta(t_2)=0, \quad (px^{\Delta\Delta})(t_3)=\int_{\eta_1}^{\eta_2}
    b(t)(px^{\Delta\Delta})(t)\nabla t,
\end{gather*}
where
\begin{equation}\label{fstar}
 f^*(t,x(t)):=f(t,x^\dagger(t)) + y(t), \quad
 x^\dagger(t):=\max\{x(t)-x^*(t),0\},
\end{equation}
such that $x^*$ given in \eqref{form} is the solution of \eqref{bvp2},
\eqref{bvpbc}, and $y$ is from (H5).

For any fixed $x\in\mathcal{P}$, $x^\dagger\le x\le \|x\|$ and by (H5),
\begin{align*}
&\int_{\rho(t_1)}^{t_3}G(t,s)f^*(s,x(s))\nabla s\\
&\le \int_{\rho(t_1)}^{t_3}G(t_2,s)\left(z(s)h(x^\dagger(s))+y(s)\right)\nabla s \\
&\le \Big(\max_{0\le\tau\le\|x\|}h(\tau)+1\Big)
\int_{\rho(t_1)}^{t_3}G(t_2,s)\left(z(s)+y(s)\right)\nabla s<\infty.
\end{align*}
For $A$ in \eqref{A} and using \eqref{AAbar}, we have
$$
A(z+y) \le \bar{A}\int_{\rho(t_1)}^{t_3}(z(s)+y(s))\nabla s;
 $$
likewise for $B$ in \eqref{B} and using (H5),
\begin{align*}
 &B(z+y)\\
& =  \frac{1}{D}\begin{vmatrix}
         \int_{\eta_1}^{\eta_2}b(t)\nabla t-1
  & \int_{\xi_1}^{\xi_2}a(t)u_2(t)\nabla t \\
    \int_{\eta_1}^{\eta_2}b(t)\big(\int_{t}^{t_3}(z(s)+y(s))\nabla s\big)
    \nabla t
  &  \int_{\xi_1}^{\xi_2}a(t)\big(\int_{\rho(t_1)}^{t_3}G(t,s)(z(s)+y(s))
     \nabla s\big)\nabla t  \end{vmatrix} \\
&\le \frac{1}{D}\begin{vmatrix}
         \int_{\eta_1}^{\eta_2}b(t)\nabla t-1 & \int_{\xi_1}^{\xi_2}a(t)u_2(t)\nabla t \\
         \int_{\eta_1}^{\eta_2}b(t)\nabla t & u_2(t_2)\int_{\xi_1}^{\xi_2}a(t)\nabla t \end{vmatrix}\int_{\rho(t_1)}^{t_3}(z(s)+y(s))\nabla s \\
& <  \infty.
\end{align*}
This allows us to define for $y\in\mathcal{P}$ the operator $T:\mathcal{P}\to \mathcal{B}$ for $t\in [\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$ by
\begin{equation}\label{T}
 (Tx)(t):=\lambda\Big(\int_{\rho(t_1)}^{t_3} G(t,s)f^*(s,x(s))\nabla s
+ A(f^*)u_2(t) +B(f^*)(u_2(t_2)-u_2(t))\Big),
\end{equation}
using \eqref{A}, \eqref{B}, and \eqref{fstar}.

\begin{lemma}\label{lemma31}
Assume {\rm (H1)} through {\rm (H5)}. Then
$T:\mathcal{P}\to\mathcal{P}$ is completely continuous.
\end{lemma}

\begin{proof}
For any $x\in\mathcal{P}$, Lemmas \ref{lemma13}, \ref{lemma15} and
Lemma \ref{lemma23} imply that $(Tx)(t)\ge\theta \|Tx\|$ on
$[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$, so that $T(\mathcal{P})\subseteq\mathcal{P}$.
By a standard application of the Arzela-Ascoli Theorem, $T$ is completely
continuous.
\end{proof}

 To establish an existence result we will employ the following fixed point
theorem due to Guo and Krasnoselskii
\cite{gu}, and seek a fixed point of $T$ in $\mathcal{P}$.

\begin{theorem}\label{fixedpt}
Let $E$ be a Banach space, $P\subseteq E$ be a cone, and suppose that
$\mathcal{S}_1$, $\mathcal{S}_2$ are bounded open balls of $E$ centered at
the origin with $\overline{\mathcal{S}}_1\subset\mathcal{S}_2$.
Suppose further that
$L:P\cap(\overline{\mathcal{S}}_2\setminus\mathcal{S}_1)\to P$
is a completely continuous
operator such that either
\begin{itemize}
 \item[$(i)$] $\|Ly\| \le \|y\|$, $y\in P\cap\partial\mathcal{S}_1$ and $\|Ly\| \ge \|y\|$,
      $y\in P\cap\partial\mathcal{S}_2$, or
 \item[$(ii)$] $\|Ly\| \ge \|y\|$, $y\in P\cap\partial\mathcal{S}_1$ and $\|Ly\| \le \|y\|$,
      $y\in P\cap\partial\mathcal{S}_2$
\end{itemize}
holds.  Then $L$ has a fixed point in
$P\cap(\overline{\mathcal{S}}_2\setminus\mathcal{S}_1)$.
\end{theorem}

\begin{theorem}\label{ethm1}
Assume {\rm (H1)} through {\rm (H5)}. Then there exists
$\lambda^*>0$ such that the third-order nonlocal time scale
boundary value problem \eqref{bvp}, \eqref{bvpbc} has at least one
positive solution in $\mathcal{P}$ for any
$\lambda\in(0,\lambda^*)$.
\end{theorem}

\begin{proof}
By Lemma \ref{lemma31}, $T:\mathcal{P}\to\mathcal{P}$ given by \eqref{T}
is completely continuous.  Take
$\mathcal{S}_1:=\{x\in\mathcal{B}:\|x\|<\Theta\}$ for $\Theta$ given
in \eqref{Theta}, and let
$$
\lambda^*:=\min\Bigg\{1,\; \frac{\int_{\rho(t_1)}^{t_3}y(s)\nabla s}{\theta
\Big(\max\limits_{0\le\tau\le\Theta}h(\tau)+1\Big)
\int_{\rho(t_1)}^{t_3}(z(s)+y(s))\nabla s}\Bigg\}.
$$
Then for any $x\in\mathcal{P}\cap\partial\mathcal{S}_1$,
$$
0 \le x^\dagger(s) \le x(s) \le \|x\| = \Theta, \quad
s\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T},
$$
and, for $\bar{A}$ as in the statement of Lemma \ref{lemma23},
\begin{align*}
 (Tx)(t) &\le \lambda\Big(\int_{\rho(t_1)}^{t_3} G(t_2,s)f^*(s,x(s))\nabla s
 +A(f^*)u_2(t_2) \Big)  \\
         &\le  \lambda \Big(\max_{0\le\tau\le\|x\|}h(\tau)+1\Big)
\int_{\rho(t_1)}^{t_3}G(t_2,s)\left(z(s)+y(s)\right)\nabla s \\
         &\quad + \lambda \bar{A}\Big(\max_{0\le\tau\le\|x\|}h(\tau)+1\Big)
\Big(\int_{\rho(t_1)}^{t_3}\big(z(s)+y(s)\big)\nabla s\Big)u_2(t_2) \\
         &\le \lambda^* u_2(t_2)\left(1+\bar{A}\right)
 \Big(\max_{0\le\tau\le\|x\|}h(\tau)+1\Big)\int_{\rho(t_1)}^{t_3}
\left(z(s)+y(s)\right)\nabla s \\
         &\le \Theta=\|x\|.
\end{align*}
Hence $\|Tx\|\le\|x\|$ for $x\in\mathcal{P}\cap\partial\mathcal{S}_1$. Pick $\Upsilon\in\mathbb{R}$ such that $\Upsilon>0$ and
$$
1 \le \frac{\lambda \Upsilon \theta}{\Theta+1}\int_{\xi_1}^{t_2}G(\xi_1,s)
\nabla s.
$$
By (H4), for any $t\in[\xi_1,t_2]_\mathbb{T}$, there exists a constant $K>0$
such that $f(t,y)>\Upsilon y$ for $y>K$.  Pick
$Q:=\max\big\{\lambda(\Theta+1),\Theta+1,\frac{K(\Theta+1)}{\theta}\big\}$.
If $\mathcal{S}_2:=\{y\in\mathcal{B}:\|y\|<Q\}$, then for any
$x\in \mathcal{P}\cap\partial\mathcal{S}_2$ and
$t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$,
\begin{align*}
 x(t)-x^*(t)
 &\ge x(t)-\lambda\theta\Theta
 \ge x(t)-\frac{\lambda\Theta}{Q}x(t)\\
 &\ge \big(1-\frac{\lambda\Theta}{Q}\big)x(t)
 \ge \big(1-\frac{\lambda\Theta}{\lambda(\Theta+1)}\big)x(t)\\
&=\frac{x(t)}{\Theta+1}\ge 0.
\end{align*}
Thus
$$
\min_{t\in[\xi_1,t_2]_\mathbb{T}}(x(t)-x^*(t))\ge\min_{t\in[\xi_1,t_2]_\mathbb{T}}
\frac{x(t)}{\Theta+1}\ge\frac{\theta Q}{\Theta+1}\ge K,
$$
so that
\begin{align*}
&\min_{t\in[\xi_1,t_2]_\mathbb{T}}(Tx)(t)\\
  & =  \min_{t\in[\xi_1,t_2]_\mathbb{T}} \lambda
\Big(\int_{\rho(t_1)}^{t_3} G(t,s)f^*(s,x(s))\nabla s+A(f^*)u_2(t)
+ B(f^*)(u_2(t_2)-u_2(t))\Big) \\
  &\ge \lambda \int_{\xi_1}^{t_2}G(\xi_1,s)f^*(s,x(s))\nabla s  \\
  &\ge \lambda \Upsilon \int_{\xi_1}^{t_2}G(\xi_1,s)(x(s)-x^*(s))\nabla s \ge \frac{\lambda \Upsilon\theta Q}{\Theta+1} \int_{\xi_1}^{t_2}G(\xi_1,s)\nabla s \\
  & =  \frac{\lambda \Upsilon\theta\|x\|}{\Theta+1}
\int_{\xi_1}^{t_2}G(\xi_1,s)\nabla s \ge \|x\|.
\end{align*}
Hence for $x\in\mathcal{P}\cap\partial\mathcal{S}_2$ we have $\|Tx\|\ge\|x\|$.
By Theorem \ref{fixedpt}, $T$ has a fixed point $x$ such that
$\Theta \le \|x\| \le Q$.  But then
$$
x(t)-x^*(t)\ge \theta\Theta-\lambda\theta\Theta
\ge (1-\lambda)\theta\Theta \ge 0.
$$
As a consequence, this $x$ solves the boundary-value problem
\begin{gather*}
   \left(px^{\Delta\Delta}\right)^\nabla(t) = \lambda (f(t,x(t)-x^*(t))+y(t)),
\quad t\in[t_1,t_3]_\mathbb{T},  \\
   \alpha x(\rho(t_1))-\beta x^\Delta(\rho(t_1))=\int_{\xi_1}^{\xi_2} a(t)x(t)
\nabla t,  \\
   x^\Delta(t_2)=0, \quad
(px^{\Delta\Delta})(t_3)=\int_{\eta_1}^{\eta_2} b(t)(px^{\Delta\Delta})(t)
\nabla t.
\end{gather*}
Now set $X(t):=x(t)-x^*(t)$ for $x^*$ given in \eqref{form}.
Then $\left(px^{\Delta\Delta}\right)^\nabla
= \left(pX^{\Delta\Delta}\right)^\nabla
+ \left(px^{*\Delta\Delta}\right)^\nabla$. As $x^*$ is the solution
of \eqref{bvp2}, \eqref{bvpbc}, we see that
\begin{gather*}
   \left(pX^{\Delta\Delta}\right)^\nabla(t) = \lambda f(t,X(t)), \quad t\in[t_1,t_3]_\mathbb{T},  \\
   \alpha X(\rho(t_1))-\beta X^\Delta(\rho(t_1))=\int_{\xi_1}^{\xi_2} a(t)X(t)\nabla t,  \\
   X^\Delta(t_2)=0, \quad \left(pX^{\Delta\Delta}\right)(t_3)=\int_{\eta_1}^{\eta_2} b(t)\left(pX^{\Delta\Delta}\right)(t)\nabla t,
\end{gather*}
in other words, $X$ is a positive solution of the third-order nonlocal
time scale boundary value problem \eqref{bvp}, \eqref{bvpbc}.
\end{proof}


As remarked in the Introduction, the results in this paper are new
for ordinary differential equations (when $\mathbb{T} = \mathbb{R}$) and for difference
equations (when $\mathbb{T} = \mathbb{Z}$).

We now provide an example to illustrate that conditions (H1)--(H5)
 are naturally satisfied.


\begin{example} \label{exa3.4} \rm
Consider for $\mathbb{T} = \mathbb{R}$ and the following choices:
$t_1 = 0$, $t_2 = 1/2$, $t_3 = 1$; $p = 1$; $\alpha = 1 = \beta$;
$f(t,x) = t + x^2$; $\xi_1 = 1/8$, $\xi_2 = 1/4$; $\eta_1 = 5/6$,
$\eta_2 = 7/8$; $a(t) = t = b(t)$. Then, for $h(x):= 1 + x^2$ with $y = 1$
and $z = 1$, the boundary value problem \eqref{bvp}, \eqref{bvpbc}
has at least one positive solution in $\mathcal{P}$ for any
 $\lambda\in(0,\lambda^*)$, where $\lambda^*\approx 0.232513$.

 With these choices, \eqref{bvp}, \eqref{bvpbc} reduces to a third-order
 BVP involving an ordinary differential equation. It is not difficult
 to verify that conditions (H1)--(H5) are satisfied. In particular,
note that
$$
\lambda^*=\min\Big\{1,\frac{5}{8(2+\Theta^2)}\Big\}
=\frac{5}{8(2+\Theta^2)}\approx 0.232513.
$$
\end{example}


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